Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘Fourier coefficient’

Final update of notes for PHY487 (condensed matter physics)

Posted by peeterjoot on January 20, 2014

Here is what will likely be the final update of my class notes from Winter 2013, University of Toronto Condensed Matter Physics course (PHY487H1F), taught by Prof. Stephen Julian.

Official course description: “Introduction to the concepts used in the modern treatment of solids. The student is assumed to be familiar with elementary quantum mechanics. Topics include: bonding in solids, crystal structures, lattice vibrations, free electron model of metals, band structure, thermal properties, magnetism and superconductivity (time permitting)”

This document contains:

• Plain old lecture notes. These mirror what was covered in class, possibly augmented with additional details.
• Personal notes exploring details that were not clear to me from the lectures, or from the texts associated with the lecture material.
• Assigned problems. Like anything else take these as is.
• Some worked problems attempted as course prep, for fun, or for test preparation, or post test reflection.
• Links to Mathematica workbooks associated with this course.
My thanks go to Professor Julian for teaching this course.

NOTE: This v.5 update of these notes is still really big (~18M).  Some of my mathematica generated 3D images result in very large pdfs.

Changelog for this update (relative to the first, and second, and third, and the last pre-exam Changelogs).

January 19, 2014 Quadratic Deybe

January 19, 2014 One atom basis phonons in 2D

January 07, 2014 Two body harmonic oscillator in 3D
Figure out a general solution for two interacting harmonic oscillators, then use the result to calculate the matrix required for a 2D two atom diamond lattice with horizontal, vertical and diagonal nearest neighbour coupling.

December 04, 2013 Lecture 24: Superconductivity (cont.)

December 04, 2013 Problem Set 10: Drude conductivity and doped semiconductors.

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Fourier coefficient integral for periodic function

Posted by peeterjoot on November 5, 2013

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In phy487 we’ve been using the fact that a periodic function

\begin{aligned}V(\mathbf{r}) = V(\mathbf{r} + \mathbf{r}_n),\end{aligned} \hspace{\stretch{1}}(1.1)

where

\begin{aligned}\mathbf{r}_n = a_1 \mathbf{a}_1 + a_2 \mathbf{a}_2 + a_3 \mathbf{a}_3,\end{aligned} \hspace{\stretch{1}}(1.2)

has a Fourier representation

\begin{aligned}V(\mathbf{r}) = \sum_\mathbf{G} V_\mathbf{G} e^{ i \mathbf{G} \cdot \mathbf{r} }.\end{aligned} \hspace{\stretch{1}}(1.3)

Here \mathbf{G} is a vector in reciprocal space, say

\begin{aligned}\mathbf{G}_{rst} = r \mathbf{g}_1 + s \mathbf{g}_2 + t \mathbf{g}_3,\end{aligned} \hspace{\stretch{1}}(1.4)

where

\begin{aligned}\mathbf{g}_i \cdot \mathbf{a}_j = 2 \pi \delta_{ij}.\end{aligned} \hspace{\stretch{1}}(1.5)

Now let’s express the explicit form for the Fourier coefficient V_\mathbf{G} so that we can compute the Fourier representation for some periodic potentials for some numerical experimentation. In particular, let’s think about what it meant to integrate over a unit cell. Suppose we have a parameterization of the points in the unit cell

\begin{aligned}\mathbf{r} = u \mathbf{a}_1 + v \mathbf{a}_2 + w \mathbf{a}_3,\end{aligned} \hspace{\stretch{1}}(1.6)

as sketched in fig. 1.1. Here u, v, w \in [0, 1]. We can compute the values of u, v, w for any vector \mathbf{r} in the cell by reciprocal projection

Fig 1.1: Unit cell

\begin{aligned}\mathbf{r} = \frac{1}{{2 \pi}} \left(  \left(  \mathbf{r} \cdot \mathbf{g}_1 \right) \mathbf{a}_1 + \left(  \mathbf{r} \cdot \mathbf{g}_2 \right) \mathbf{a}_2 + \left(  \mathbf{r} \cdot \mathbf{g}_3 \right) \mathbf{a}_3 \right),\end{aligned} \hspace{\stretch{1}}(1.7)

or

\begin{aligned}\begin{aligned}u(\mathbf{r}) &= \frac{1}{{2 \pi}} \mathbf{r} \cdot \mathbf{g}_1 \\ v(\mathbf{r}) &= \frac{1}{{2 \pi}} \mathbf{r} \cdot \mathbf{g}_2 \\ w(\mathbf{r}) &= \frac{1}{{2 \pi}} \mathbf{r} \cdot \mathbf{g}_3.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.8)

Let’s suppose that \mathbf{V}(\mathbf{r}) is period in the unit cell spanned by \mathbf{r} = u \mathbf{a}_1 + v \mathbf{a}_2 + w \mathbf{a}_3 with u, v, w \in [0, 1], and integrate over the unit cube for that parameterization to compute V_\mathbf{G}

\begin{aligned}\int_0^1 du\int_0^1 dv\int_0^1 dwV( u \mathbf{a}_1 + v \mathbf{a}_2 + w \mathbf{a}_3 ) e^{-i \mathbf{G}' \cdot \mathbf{r} }=\sum_{r s t}V_{\mathbf{G}_{r s t}}\int_0^1 du\int_0^1 dv\int_0^1 dwe^{-i \mathbf{G}' \cdot \mathbf{r} }e^{i \mathbf{G} \cdot \mathbf{r} }\end{aligned} \hspace{\stretch{1}}(1.9)

Let’s write

\begin{aligned}\begin{aligned}\mathbf{G} &= r \mathbf{g}_1 + s \mathbf{g}_2 + t \mathbf{g}_3 \\ \mathbf{G} &= r' \mathbf{g}_1 + s' \mathbf{g}_2 + t' \mathbf{g}_3,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.10)

so that

\begin{aligned}e^{-i \mathbf{G}' \cdot \mathbf{r} } e^{i \mathbf{G} \cdot \mathbf{r} }=e^{ 2 \pi i (r - r') u } e^{ 2 \pi i (s - s') u } e^{ 2 \pi i (t - t') u } \end{aligned} \hspace{\stretch{1}}(1.11)

Picking the u integral of this integrand as representative, we have when r = r'

\begin{aligned}\int_0^1 du e^{ 2 \pi i (r - r') u } =\int_0^1 du= 1,\end{aligned} \hspace{\stretch{1}}(1.12)

and when r \ne r'

\begin{aligned}\int_0^1 du e^{ 2 \pi i (r - r') u } ={\left.{{   \frac{    e^{ 2 \pi i (r - r') u }   }   {   2 \pi i (r - r')    }}}\right\vert}_{{u = 0}}^{{1}}=\frac{1}{{2 \pi i (r - r') }} \left(  e^{ 2 \pi i (r - r') } - 1  \right).\end{aligned} \hspace{\stretch{1}}(1.13)

This is just zero since r - r' is an integer, so we have

\begin{aligned}\int_0^1 du e^{ 2 \pi i (r - r') u } = \delta_{r, r'}.\end{aligned} \hspace{\stretch{1}}(1.14)

This gives us

\begin{aligned}\int_0^1 du\int_0^1 dv\int_0^1 dwV( u \mathbf{a}_1 + v \mathbf{a}_2 + w \mathbf{a}_3 ) e^{ -2 \pi i r' u } e^{ -2 \pi i s' v } e^{ -2 \pi i t' w } =\sum_{r s t}V_{\mathbf{G}_{r s t}}\delta_{r s t, r' s' t'}= V_{\mathbf{G}_{r' s' t'}}.\end{aligned} \hspace{\stretch{1}}(1.15)

This is our \textAndIndex{Fourier coefficient}. The \textAndIndex{Fourier series} written out in gory but explicit detail is

\begin{aligned}\boxed{V( u \mathbf{a}_1 + v \mathbf{a}_2 + w \mathbf{a}_3 ) = \sum_{r s t}\left(  \int_0^1 du' \int_0^1 dv' \int_0^1 dw' V( u' \mathbf{a}_1 + v' \mathbf{a}_2 + w' \mathbf{a}_3 ) e^{ -2 \pi i (r u' + s v' + t w') }  \right)e^{ 2 \pi i (r u + s v + t w) }.}\end{aligned} \hspace{\stretch{1}}(1.16)

Also observe the unfortunate detail that we require integrability of the potential in the unit cell for the Fourier integrals to converge. This prohibits the use of the most obvious potential for numerical experimentation, the inverse radial V(\mathbf{r}) = -1/\left\lvert {\mathbf{r}} \right\rvert.

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A Fourier series refresher.

Posted by peeterjoot on May 3, 2012

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Motivation.

I’d used the wrong scaling in a Fourier series over a [0, 1] interval. Here’s a reminder to self what the right way to do this is.

Guts

Suppose we have a function that is defined in terms of a trigonometric Fourier sum

\begin{aligned}\phi(x) = \sum c_k e^{i \omega k x},\end{aligned} \hspace{\stretch{1}}(2.1)

where the domain of interest is x \in [a, b]. Stating the problem this way avoids any issue of existence. We know c_k exists, but just want to find what they are given some other representation of the function.

Multiplying and integrating over our domain we have

\begin{aligned}\begin{aligned}\int_a^b \phi(x) e^{-i \omega m x} dx &= \sum c_k \int_a^b e^{i \omega (k -m) x} dx \\ &= c_m (b - a) + \sum_{k \ne m} \frac{e^{i \omega(k-m) b} - e^{i \omega(k-m)a}}{i \omega (k -m)} .\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.2)

We want all the terms in the sum to be be zero, requiring equality of the exponentials, or

\begin{aligned}e^{i \omega (k -m) (b -a )} = 1,\end{aligned} \hspace{\stretch{1}}(2.3)

or

\begin{aligned}\omega = \frac{2 \pi}{b - a}.\end{aligned} \hspace{\stretch{1}}(2.4)

This fixes our Fourier coefficients

\begin{aligned}c_m = \frac{1}{{b - a}} \int_a^b \phi(x) e^{- 2 \pi i m x/(b - a)} dx.\end{aligned} \hspace{\stretch{1}}(2.5)

Given this, the correct (but unnormalized) Fourier basis for a [0, 1] interval would be the functions e^{2 \pi i x}, or the sine and cosine equivalents.

References

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